4

Prove: a set of vectors $K$ is linearly dependent iff a vector is linear combination of the others.

Let: $\alpha_1k_1 + \alpha_2k_2+...\alpha_nk_n = 0$

Then, There must be $\alpha_i \ne 0$. Therefore,

$$\alpha_1k_1 + \alpha_2k_2 + \alpha_{i-1}k_{i-1} + \alpha_{i+1}k_{i+1}+...+ \alpha_nk_n = -\alpha_ik_i.$$

Then,
$${\alpha_1 \over -\alpha_i}k_1 + {\alpha_2 \over -\alpha_i}k_2+...+{\alpha_n \over -\alpha_i}k_n = k_i$$

Indeed, $k_i$ is a linear combination of the other vectors.

My question is:
Is that answering the iff condition?

SuperStamp
  • 993
  • 1
    This answers the if and not the only if. –  Jan 17 '14 at 22:30
  • 1
    You need to show that if a vector is a linear combination of the others, then the collection is linearly dependent. – copper.hat Jan 17 '14 at 22:34
  • 1
    @copper.hat, isn't it just reading the current proof from bottom to top? – SuperStamp Jan 17 '14 at 22:35
  • 1
    @SuperStamp Yes, just read backwards. All your implications go both ways. – T.J. Gaffney Jan 17 '14 at 22:36
  • Two questions: (1) is it a rigorous proof, I if I write "read backward to prove other direction"? (2) Instead, I could prove the only if as @SamiBenRomdhane mentioned. Which means: if the set is linearly independent then non of the vector is linear combination of the others. Right? – SuperStamp Jan 17 '14 at 22:40
  • 1
    @SuperStamp: It depends on your audience. Instructing your reader to read the steps in reverse can work, but it is completely unambiguous and less commanding to just establish a sequence of equivalences. – copper.hat Jan 17 '14 at 22:43
  • @SuperStamp: formally, the reverse direction is no valid proof, but it is very easy to transform it into one. – Doc Brown Jan 17 '14 at 22:49
  • BTW, the above can easily be reinforced a little as follows: a well-ordered set of vectors (for example, a finite or countable infinite one) is linearly dependent iff there is one vector linearly dependent on the preceeding ones. This can be pretty handy when dealing with some exercises. – DonAntonio Jan 17 '14 at 22:51
  • Your proof, as written, is incorrect. Why do you have

    $${\alpha_1 \over -\alpha_1}k_1 + {\alpha_2 \over -\alpha_2}k_2+\cdots+{\alpha_n \over -\alpha_n}k_n = k_i$$

    It is very rare that

    $$-k_1 -k_2-\cdots -k_n = k_i$$

    Perhaps your mean

    $${\alpha_1 \over -\alpha_i}k_1 + {\alpha_2 \over -\alpha_i}k_2+\cdots+{\alpha_n \over -\alpha_i}k_n = k_i$$

     – Fly by Night Jan 17 '14 at 22:51
  • @FlybyNight, it was a typo(s). I'll correct that – SuperStamp Jan 17 '14 at 22:55

1 Answers1

4

Assume that there exist $a_i \in \mathbb{K}$ for which $a_1{\bf v}_1+\cdots+a_n{\bf v}_n = {\bf 0}$, where not all $a_i$ are zero.

Then, as you say, there exists an $a_k \neq 0$, and we can write

$${\bf v}_k = -\frac{1}{a_k}(a_1{\bf v}_1+\cdots + a_{k-1}{\bf v}_{k-1}+a_{k+1}{\bf v}_{k+1}+\cdots+a_n{\bf v}_n)$$

Hence ${\bf v}_k$ is a linear combination of the other ${\bf v}_i$.

Assume that ${\bf v}_k$ is a linear combination of the other ${\bf v}_i$. In that case, there exist $\lambda_j \in \mathbb{K}$ for which $${\bf v}_k = \lambda_1{\bf v}_1 + \cdots + \lambda_{k-1}{\bf v}_{k-1}+\lambda_{k+1}{\bf v}_{k+1}+\cdots + \lambda_n{\bf v}_n$$

Putting ${\bf v}_k=\sum_{i \neq k} \lambda_i{\bf v}_i$ into $a_1{\bf v}_1+\cdots + a_n{\bf v}_n$ shows that $\lambda_j = -a_j$ gives

$$a_1{\bf v}_1+\cdots+a_n{\bf v}_n = {\bf 0}$$

Fly by Night
  • 32,272